[alex-g]

At what sort of age do you think people are ready for proof by contradiction? I remember being taught it explicitly at secondary school (in the UK) but we may have seen implicit uses of it earlier. I would imagine that some very young children might find it difficult to remember the train of thought.

[ColinWright]

That's why flipping it to say that every rational, when squared, doesn't give you a prime. The concept of irrartional is already tough. I've found that when handled carefully, even quite young kids can handle at least some of it.

But you're right, proof by contradiction can be tough.

[tokenadult]

To answer your question, the youngest age at which I have seen a pupil propose a proof by contradiction to me was age eight. The girl had read some Life of Fred

http://www.lifeoffredmath.com/

books for children about mathematics, and had newly joined my local mathematics class. On only the second or third week of class, she came up to me after class and said, "I've discovered a proof by contradiction for the parallel postulate." As you can imagine, I found this quite amazing. (I knew her mother, and thus knew the daughter a little before she joined my class, but I would say that's rather precocious behavior even in the social circle I keep.) Her "proof," of course, was really Saccheri's flawed proof

http://www.jimloy.com/geometry/saccheri.htm

http://www-history.mcs.st-and.ac.uk/HistTopics/Non-Euclidean...

that assumed the postulate to show the "impossibility" of any quadrilateral that didn't fit Euclidean geometry. But most of us have minds that begin study of mathematics with a stubbornly Euclidean set of presuppositions, so that was all right. The girl eventually advanced from my mathematics class to my colleague's more advanced class, and then did a summer at Epsilon Camp

http://www.epsiloncamp.org/

in that program's first year of existence.

[trhtrsh]

Proof by contradiction is almost never necessary.

Proof that if a number has a rational number square root, it must be a square:

    Let's suppose that sqrt(p) = a/b
    That means that p = (a^2)/(b^2)
    That means that p * b^2 = a^2

Therefore p is a square (by counting prime factors).